## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 22 - Problems - Page 863: 24

#### Answer

The only combination of $\epsilon_0$ and $\mu_0$ with the dimensions of speed is $(\epsilon_0~\mu_0)^{-\frac{1}{2}}$

#### Work Step by Step

$\epsilon_0 = 8.85\times 10^{-12}~m^{-3}~kg^{-1}~s^4~A^2$ $\mu_0 = 1.26\times 10^{-6}~m~kg~s^{-2}~A^{-2}$ Speed $v$ has units of m/s. Suppose $v = C~\epsilon_0^a~\mu_0^b$, where $C$ is some dimensionless constant. Consider the units of kilograms: $(kg^{-1})^a~(kg)^b = kg^0$ $b-a = 0$ $a = b$ Consider the units of meters: $(m^{-3})^a~(m)^b = m^1$ $b - 3a = 1$ $a - 3a = 1$ $a = -\frac{1}{2}$ Then $b = -\frac{1}{2}$, since $a = b$ $v = C~\epsilon_0^a~\mu_0^b$ $v = C~\epsilon_0^{-\frac{1}{2}}~\mu_0^{-\frac{1}{2}}$ $v = C~(\epsilon_0~\mu_0)^{-\frac{1}{2}}$ The only combination of $\epsilon_0$ and $\mu_0$ with the dimensions of speed is $(\epsilon_0~\mu_0)^{-\frac{1}{2}}$

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